Đề HSG 11

Inequalities from 2008 Mathematical Competition      Inequalities from 2008 Mathematical Competition Editor Manh Dung Nguyen, High School for Gifted Students, HUS, Vietnam Contact If you have any question about this ebook, please contact us. Email: nguyendunghus@gmail.com Acknowledgments We thank a lot to Mathlinks Forum 1 and their members for the reference to problems and many nice solutions from them! Hanoi, 10 October 2008 1 Website: http://mathlinks.ro Inequalities from 2008 Mathematical Competition      Abbreviations • IMO International mathematical Olympiad • TST Team Selection Test • MO Mathematical Olympiad • LHS Left hand side • RHS Right hand side • W.L.O.G Without loss of generality •  :  cyclic Contents 1 Problems 4 2 Solutions 10 3 Chapter 1 Problems Pro 1. (Vietnamese National Olympiad 2008) Let x, y, z be distinct non-negative real numbers. Prove that 1 (x − y) 2 + 1 (y − z) 2 + 1 (z − x) 2 ≥ 4 xy + yz + zx . ∇ Pro 2. (Iranian National Olympiad (3rd Round) 2008). Find the smallest real K such that for each x, y, z ∈ R + : x √ y + y √ z + z √ x ≤ K  (x + y)(y + z)(z + x) ∇ Pro 3. (Iranian National Olympiad (3rd Round) 2008). Let x, y, z ∈ R + and x + y + z = 3. Prove that: x 3 y 3 + 8 + y 3 z 3 + 8 + z 3 x 3 + 8 ≥ 1 9 + 2 27 (xy + xz + yz) ∇ Pro 4. (Iran TST 2008.) Let a, b, c > 0 and ab +ac +bc = 1. Prove that: √ a 3 + a + √ b 3 + b + √ c 3 + c ≥ 2 √ a + b + c ∇ 4 Inequalities from 2008 Mathematical Competition      Pro 5. (Macedonian Mathematical Olympiad 2008.) Positive num- bers a, b, c are such that (a + b) (b + c) (c + a) = 8. Prove the inequality a + b + c 3 ≥ 27  a 3 + b 3 + c 3 3 ∇ Pro 6. (Mongolian TST 2008) Find the maximum number C such that for any nonnegative x, y, z the inequality x 3 + y 3 + z 3 + C(xy 2 + yz 2 + zx 2 ) ≥ (C + 1)(x 2 y + y 2 z + z 2 x). holds. ∇ Pro 7. (Federation of Bosnia, 1. Grades 2008.) For arbitrary reals x, y and z prove the following inequality: x 2 + y 2 + z 2 − xy − yz − zx ≥ max{ 3(x − y) 2 4 , 3(y − z) 2 4 , 3(y − z) 2 4 }. ∇ Pro 8. (Federation of Bosnia, 1. Grades 2008.) If a, b and c are positive reals such that a 2 + b 2 + c 2 = 1 prove the inequality: a 5 + b 5 ab(a + b) + b 5 + c 5 bc(b + c) + c 5 + a 5 ca(a + b) ≥ 3(ab + bc + ca) − 2 ∇ Pro 9. (Federation of Bosnia, 1. Grades 2008.) If a, b and c are positive reals prove inequality: (1 + 4a b + c )(1 + 4b a + c )(1 + 4c a + b ) > 25 ∇ Pro 10. (Croatian Team Selection Test 2008) Let x, y, z be positive numbers. Find the minimum value of: (a) x 2 + y 2 + z 2 xy + yz (b) x 2 + y 2 + 2z 2 xy + yz Inequalities from 2008 Mathematical Competition      ∇ Pro 11. (Moldova 2008 IMO-BMO Second TST Problem 2) Let a 1 , . . . , a n be positive reals so that a 1 + a 2 + . . . + a n ≤ n 2 . Find the minimal value of A =  a 2 1 + 1 a 2 2 +  a 2 2 + 1 a 2 3 + . . . +  a 2 n + 1 a 2 1 ∇ Pro 12. (RMO 2008, Grade 8, Problem 3) Let a, b ∈ [0, 1]. Prove that 1 1 + a + b ≤ 1 − a + b 2 + ab 3 . ∇ Pro 13. (Romanian TST 2 2008, Problem 1) Let n ≥ 3 be an odd integer. Determine the maximum value of  |x 1 − x 2 | +  |x 2 − x 3 | + . . . +  |x n−1 − x n | +  |x n − x 1 |, where x i are positive real numbers from the interval [0, 1] ∇ Pro 14. (Romania Junior TST Day 3 Problem 2 2008) Let a, b, c be positive reals with ab + bc + ca = 3. Prove that: 1 1 + a 2 (b + c) + 1 1 + b 2 (a + c) + 1 1 + c 2 (b + a) ≤ 1 abc . ∇ Pro 15. (Romanian Junior TST Day 4 Problem 4 2008) Determine the maximum possible real value of the number k, such that (a + b + c)  1 a + b + 1 c + b + 1 a + c − k  ≥ k for all real numbers a, b, c ≥ 0 with a + b + c = ab + bc + ca. ∇ Inequalities from 2008 Mathematical Competition      Pro 16. (Serbian National Olympiad 2008) Let a, b, c be positive real numbers such that x + y + z = 1. Prove inequality: 1 yz + x + 1 x + 1 xz + y + 1 y + 1 xy + z + 1 z ≤ 27 31 . ∇ Pro 17. (Canadian Mathematical Olympiad 2008) Let a, b, c be positive real numbers for which a + b + c = 1. Prove that a − bc a + bc + b − ca b + ca + c − ab c + ab ≤ 3 2 . ∇ Pro 18. (German DEMO 2008) Find the smallest constant C such that for all real x, y 1 + (x + y) 2 ≤ C · (1 + x 2 ) · (1 + y 2 ) holds. ∇ Pro 19. (Irish Mathematical Olympiad 2008) For positive real num- bers a, b, c and d such that a 2 + b 2 + c 2 + d 2 = 1 prove that a 2 b 2 cd + +ab 2 c 2 d + abc 2 d 2 + a 2 bcd 2 + a 2 bc 2 d + ab 2 cd 2 ≤ 3/32, and determine the cases of equality. ∇ Pro 20. (Greek national mathematical olympiad 2008, P1) For the positive integers a 1 , a 2 , , a n prove that   n i=1 a 2 i  n i=1 a i  kn t ≥ n  i=1 a i where k = max {a 1 , a 2 , , a n } and t = min {a 1 , a 2 , , a n }. When does the equality hold? ∇ Inequalities from 2008 Mathematical Competition      Pro 21. (Greek national mathematical olympiad 2008, P2) If x, y, z are positive real numbers with x, y, z < 2 and x 2 + y 2 + z 2 = 3 prove that 3 2 < 1 + y 2 x + 2 + 1 + z 2 y + 2 + 1 + x 2 z + 2 < 3 ∇ Pro 22. (Moldova National Olympiad 2008) Positive real numbers a, b, c satisfy inequality a + b + c ≤ 3 2 . Find the smallest possible value for: S = abc + 1 abc ∇ Pro 23. (British MO 2008) Find the minimum of x 2 + y 2 + z 2 where x, y, z ∈ R and satisfy x 3 + y 3 + z 3 − 3xyz = 1 ∇ Pro 24. (Zhautykov Olympiad, Kazakhstan 2008, Question 6) Let a, b, c be positive integers for which abc = 1. Prove that  1 b(a + b) ≥ 3 2 . ∇ Pro 25. (Ukraine National Olympiad 2008, P1) Let x, y and z are non-negative numbers such that x 2 + y 2 + z 2 = 3. Prove that: x  x 2 + y + z + y  x + y 2 + z + z  x + y + z 2 ≤ √ 3 ∇ Pro 26. (Ukraine National Olympiad 2008, P2) For positive a, b, c, d prove that (a + b)(b + c)(c + d)(d + a)(1 + 4 √ abcd) 4 ≥ 16abcd(1 + a)(1 + b)(1 + c)(1 + d) ∇ Inequalities from 2008 Mathematical Competition      Pro 27. (Polish MO 2008, Pro 5) Show that for all nonnegative real values an inequality occurs: 4( √ a 3 b 3 + √ b 3 c 3 + √ c 3 a 3 ) ≤ 4c 3 + (a + b) 3 . ∇ Pro 28. (Chinese TST 2008 P5) For two given positive integers m, n > 1, let a ij (i = 1, 2, ··· , n, j = 1, 2, ··· , m) be nonnegative real numbers, not all zero, find the maximum and the minimum values of f, where f = n  n i=1 (  m j=1 a ij ) 2 + m  m j=1 (  n i=1 a ij ) 2 (  n i=1  m j=1 a ij ) 2 + mn  n i=1  m i=j a 2 ij ∇ Pro 29. (Chinese TST 2008 P6) Find the maximal constant M, such that for arbitrary integer n ≥ 3, there exist two sequences of positive real number a 1 , a 2 , ··· , a n , and b 1 , b 2 , ··· , b n , satisfying (1):  n k=1 b k = 1, 2b k ≥ b k−1 + b k+1 , k = 2, 3, ··· , n −1; (2):a 2 k ≤ 1 +  k i=1 a i b i , k = 1, 2, 3, ··· , n, a n ≡ M. Chapter 2 Solutions Problem 1. (Vietnamese National Olympiad 2008) Let x, y, z be dis- tinct non-negative real numbers. Prove that 1 (x − y) 2 + 1 (y − z) 2 + 1 (z − x) 2 ≥ 4 xy + yz + zx . Proof. (Posted by Vo Thanh Van). Assuming z = min{x, y, z}. We have (x − z) 2 + (y − z) 2 = (x − y) 2 + 2(x − z)(y − z) So by the AM-GM inequality, we get 1 (x − y) 2 + 1 (y − z) 2 + 1 (z − x) 2 = 1 (x − y) 2 + (x − y) 2 (y − z) 2 (z − x) 2 + 2 (x − z)(y − z) ≥ 2 (x − z)(y − z) + 2 (x − z)(y − z) = 4 (x − z)(y − z) ≥ 4 xy + yz + zx We are done. Proof. (Posted by Altheman). Let f(x, y, z) denote the LHS minus the RHS. Then f(x + d, y + d, z + d) is increasing in d so we can set the least of x + d, y + d, z + d equal to zero (WLOG z = 0). Then we have 1 (x − y) 2 + 1 x 2 + 1 y 2 − 4 xy = (x 2 + y 2 − 3xy) 2 x 2 y 2 (x − y) 2 ≥ 0 ∇ 10 [...]... the inequality It becomes x3 + y 3 + cx2 y ≥ (c + 1)xy 2 Thus, we have to find the minimal value of f (y) = 1 y3 − y2 + 1 1 =y+ 2−y y y(y − 1) You can see here: http://www.mathlinks.ro/viewtopic.php?p =113 0901 Inequalities from 2008 Mathematical Competition when y > 1 It is easy to find that f (y) = 0 ⇔ 2y − 1 = (y(y − 1))2 ⇔ y 4 − 2y 3 + y 2 − 2y + 1 = 0 Solving this symmetric equation gives us: √ 1... value is 8/3 Expanding x− we get x2 + y 2 + 2z 2 − 2 y 3 2 8/3xy − + √ 1 y − 6z 3 2 ≥ 0, 8/3yz ≥ 0, so x2 + y 2 + z 2 ≥ xy + yz Equality occurs, for example, if x = 2, y = 8 3 √ 6, and z = 1 Problem 11 (Moldova 2008 IMO-BMO Second TST Problem 2) Let a1 , , an be positive reals so that a1 +a2 + .+an ≤ n Find the minimal 2 value of 1 1 1 A = a2 + 2 + a2 + 2 + + a2 + 2 1 2 n a2 a3 a1 Proof (Posted... where n f= ( n i=1 ( n i=1 m 2 j=1 aij ) m 2 j=1 aij ) +m + mn m j=1 ( n i=1 n 2 i=1 aij ) m 2 i=j aij Proof (Posted by tanpham) We will prove that the maximum value of f is 1 • For n = m = 2 Setting a11 = a, a21 = b, a12 = x, a21 = y We have 2 ((a + b)2 + (x + y)2 + (a + x)2 + (b + y)2 ) f= ≤1 (a + b + x + y)2 + 4 (a2 + b2 + x2 + y 2 ) ⇔= (x + b − a − y)2 ≥ 0 as needed • For n = 2, m = 3 Using the . yz (b) x 2 + y 2 + 2z 2 xy + yz Inequalities from 2008 Mathematical Competition      ∇ Pro 11. (Moldova 2008 IMO-BMO Second TST Problem 2) Let a 1 , . . . , a n be positive reals so that. = y 3 − y 2 + 1 y 2 − y = y + 1 y(y − 1) 1 You can see here: http://www.mathlinks.ro/viewtopic.php?p =113 0901 Inequalities from 2008 Mathematical Competition      when y > 1. It is easy to find. y 2 + z 2 xy + yz ≥  8 3 . Equality occurs, for example, if x = 2, y = √ 6, and z = 1. ∇ Problem 11. (Moldova 2008 IMO-BMO Second TST Problem 2) Let a 1 , . . . , a n be positive reals so that